Two different polynomial equations of degree $4$

Question

Let $f(x)$ and $g(x)$ are polynomial of degree 4 such that $g(\alpha) = g'(\alpha) = g''(\alpha) = 0$. If $\lim_{x \to \alpha } \frac{f( x )}{g( x)} = 0$ , then the number of different real solutions of equation $f(x)\cdot g'(x) + g(x)\cdot f'(x) = 0$ is equal to_______________

My approach is as follow

$T\left( x \right) = f\left( x \right)g'\left( x \right) + f'\left( x \right)g\left( x \right) = \frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = 0$

$f\left( x \right)g\left( x \right) = C$

$f\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d\,\& \,g\left( x \right) = {x^4} + a'{x^3} + b'{x^2} + c'x + d'$

$g\left( \alpha \right) = {\alpha ^4} + a'{\alpha ^3} + b'{\alpha ^2} + c'\alpha + d' = 0;g'\left( \alpha \right) = 4{\alpha ^3} + 3a'{\alpha ^2} + 2b'\alpha + c' = 0;g''\left( \alpha \right) = 12{\alpha ^2} + 6a'\alpha + 2b' = 0$

We have to presume that $f\left( \alpha \right) = f'\left( \alpha \right) = f''\left( \alpha \right) = f'''\left( \alpha \right) = 0$ and $g'''\left( \alpha \right) \ne 0$

$f'''\left( \alpha \right) = 36\alpha + 6a = 0 \Rightarrow a = - 6\alpha $

How we will proceed from here

Hint: Clearly $g(x)=A(x-\alpha)^3(x-\beta)$ and $f$ is ... — user10354138, Aug 21 '21 at 10:06
@anomaly There is really no improvement with putting the number 4 into MathJaX in the title. In fact I'd argue it is deterimental in this case. See old meta guideline discussion, in particular, #3 in the accepted community wiki answer there. — user10354138, Aug 21 '21 at 11:42
@user10354138: I'd say it doesn't matter one way or the other, but I did want to change the all-caps in the title. — anomaly, Aug 21 '21 at 15:21
@anomaly But the all caps has already been changed by Jochen — user10354138, Aug 21 '21 at 15:22
@user10354138: Hmm, might have been some weirdness with the editing queue. When I was changed it, it was confirming someone else's edit (I don't remember whose) rather than editing from scratch. — anomaly, Aug 21 '21 at 15:24

score 0 · Accepted Answer · answered Aug 21 '21 at 11:21

As already mentioned, from $g^{(i)}(\alpha)=0$ for $0\leq \alpha <3$ follows, that $\alpha$ is a root of order 3, i.e. $(x-\alpha)^3\mid g$. (One idea to prove this is to use the taylor expansion $g(x)=\sum_{i=0}^4\frac{g^{(i)}(\alpha)}{i!}(x-\alpha)^i$.) Since $\deg(g)=4$, there is a polynomial $p$ of degree 1 with $g(x)=(x-\alpha)^3p(x)$. As user10354138 has written, there are some constants $A\neq 0,\beta$ with $g(x)=A(x-\alpha)^3(x-\beta)$. From $\lim_{x\to\alpha}\frac{f(x)}{g(x)}=0$ follows, that $(x-\alpha)^4\mid f$. Hence, there is a constant $B\neq0$ with $f(x)=B(x-\alpha)^4$. Let $C=A\cdot B\neq 0$, then $$f(x)\cdot g(x)=C(x-\alpha)^7(x-\beta).$$ I think, you can continue from here.

Two different polynomial equations of degree $4$

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